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3 years ago

Explain how the perfume atomizer works under Bernoulli's principle

Physics

1 answer:

ss7ja [257]3 years ago

4 0

Answer:

In an atomizer, or perfume sprayer, you squeeze a rubber bulb to squirt air through a tube. Because of the Bernoulli principle, the air rushing through the tube has a lower pressure than the surrounding atmosphere. ... The perfume is pushed out of the tube and sprays into the air as a fine mist.

Explanation:

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The manipulation of natural sounds via the medium of magnetic tape is called

alexandr1967 [171]

<span>The manipulation of natural sounds via the medium of magnetic tape is called "</span>Musique concrete".

Musique concrete refers to an experimental method of melodic composition utilizing recorded sounds as crude material. The strategy was created around 1948 by the French composer Pierre Schaeffer and his partners at the Studio d'Essai ("Experimental Studio") of the French radio framework. The major guideline of musique concrète lies in the collection of different regular sounds recorded on tape (or, initially, on plates) to deliver a montage of sound.

3 0

3 years ago

An object moving on a horizontal, frictionless surface makes a glancing collision with another object initially at rest on the s

cluponka [151]

Answer:

Momentum is always conserved, and kinetic energy may be conserved.

Explanation:

For an object moving on a horizontal, frictionless surface which makes a glancing collision with another object initially at rest on the surface, the type of collision experienced by this objects can either be elastic or an inelastic collision depending on whether the object sticks together after collision or separates and move with a common velocity after collision.

If the body separates and move with a common velocity after collision, the collision is elastic but if they sticks together after collision, the collision is inelastic.

Either ways the momentum of the bodies are always conserved since they will always move with a common velocity after collision but their kinetic energy may or may not be conserved after collision, it all depends whether they separates or stick together after collision and since we are not told in question whether or not they separate, we can conclude that their kinetic energy "may" be conserved.

6 0

2 years ago

An important diagnostic tool for heart disease is the pressure difference between blood pressure in the heart and in the aorta l

butalik [34]

Answer:

a) f ’’ = f₀ $\frac{1 + \frac{v}{c} }{1- \frac{v}{c} }$ , b) Δf = 2 f₀ $\frac{v}{c}$

Explanation:

a) This is a Doppler effect exercise, which we must solve in two parts in the first the emitter is fixed and in the second when the sound is reflected the emitter is mobile.

Let's look for the frequency (f ’) that the mobile aorta receives, the blood is leaving the aorta or is moving towards the source

f ’= fo $\frac{c+v}{c}$

This sound wave is reflected by the blood that becomes the emitter, mobile and the receiver is fixed.

f ’’ = f’ $\frac{c}{ c-v}$

where c represents the sound velocity in stationary blood

therefore the received frequency is

f ’’ = f₀ $\frac{c}{c-v}$

let's simplify the expression

f ’’ = f₀ \frac{c+v}{c-v}

f ’’ = f₀ $\frac{1 + \frac{v}{c} }{1- \frac{v}{c} }$

b) At the low speed limit v <c, we can expand the quantity

(1 -x)ⁿ = 1 - x + n (n-1) x² + ...

$( 1- \frac{v}{c} ) ^{-1} = 1 + \frac{v}{c}$

f ’’ = fo $( 1+ \frac{v}{c}) ( 1 + \frac{v}{c} )$

f ’’ = fo $( 1 + 2 \frac{v}{c} + \frac{v^2}{ c^2} )$

leave the linear term

f ’’ = f₀ + f₀ 2 $\frac{v}{c}$

the sound difference

f ’’ -f₀ = 2f₀ v/c

Δf = 2 f₀ $\frac{v}{c}$

4 0

2 years ago

An object of mass 10.0kg is released at point A, slidesto the bottom of the 30° incline, then collides with ahorizontal massless

Elenna [48]

Answer:

Explanation:

Potential energy lost by mass = mgh

= 10 x 9.8 x 2 = 196 J

a ) If v be velocity of mass at the bottom , its kinetic energy will be stored in spring as elastic energy

= 1/2 m v² = 1/2 k x² , k is spring constant , x is compression , m is mass falling down

.5 x 10 v² = .5 x 500 x .75²

v = 5.3 m /s

b ) kinetic energy of mass at the bottom

= /2 m v²

= .5 x 10 x 5.3²

= 140.45 J

energy lost by mass while coming down

=potential energy at the top - kinetic energy at bottom

= 196 - 140.45

= 55.55 J .

This is equal to negative work done by friction

work done by friction = - 55.55 J

c ) Since there will be no loss of energy in compression and extension of spring so , no loss of kinetic energy will take place of mass . So it wil have same velocity that is 5.3 m /s while on its return journey.

d ) kinetic energy at the bottom = 140.45

loss of energy by friction again

= 140.45 - 55.55

= 84.9 J

If h be the height attained

mgh = 84.9

10 x 9.8 x h = 84.9

h = .866 m

( We have assumed that loss of energy in return journey will be same due to friction . )

6 0

3 years ago

A copper sphere was moving at 25 m/s when it hit another object. This caused all of the KE to be converted into thermal energy f

JulsSmile [24]

Increased temperature is 0.81 c

7 0

3 years ago